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प्रश्न
A monkey is sitting on a tree limb. The limb exerts a normal force of 48 N and a frictional force of 20 N. Find the magnitude of the total force exerted by the limb on the monkey.
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उत्तर
Given: The limb of the tree exerts a normal force of 48 N and a frictional force of 20 N.
So, resultant magnitude of the force if given by
\[R = \sqrt{\left( {48}^2 + {20}^2 \right)}\]
\[ = \sqrt{2304 + 400}\]
\[ = \sqrt{2704} = 52 N\]
∴ The magnitude of the total force exerted by the limb on the monkey is 52 N.
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